Simple physics (Final) (Pic provided)

when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0

3 years ago

A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?

seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

The object describes a curved path ruled exclusively by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the following equation:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Note it doesn't depend on the initial velocity but on the height.

The bullet is fired horizontally at h=1.3 m, thus:

$\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}$

$\displaystyle t=\sqrt{\frac{2.6}{9.8}}$

t = 0.52 s

The bullet was 0.52 seconds in the air.

3 0

2 years ago

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

4 0

3 years ago

A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0

2 years ago

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ????????,2=17.9vi,2=17.9

AleksAgata [21]

Answer:

V=27.24 m/s

Explanation:

We need to apply the linear momentum conservation theorem:

$m_1*v_{o1}+m_2*v_{o2}=m_t*v_{f}\\$

The velocity of the eagle its defined by its two components:

$v_x=V*cos(\theta)\\v_y=V*sin(\theta)\\v_x=35.9*cos(61.9^o)=16.9m/s\\v_y=35.9*sin(61.9^o)=31.7m/s$

$2*(16.9m/s(i)+31.66m/s(j))+17.9m/s(i)=3*v_f\\v_f=17.23m/s(i)+21.10m/s(j)$

because speed is a scalar value:

$V=\sqrt{(21.10m/s)^2+(17.23)^2}\\V=27.24m/s$

6 0

3 years ago